A 3.9-kg block is attached to a spring with a force constant of 565 N/m, as show

Question

A 3.9-kg block is attached to a spring with a force constant of 565 N/m, as shown in the figure.

(a) Find the work done by the spring on the block as the block moves from A to B along paths 1 and 2.

2) Pushing on the pump of a soap dispenser compresses a small spring. When the spring is compressed 0.48 cm, its potential energy is 0.0025 J.

(a) What is the force constant of the spring?

(b) What compression is required for the spring potential energy to equal 0.0078 J?

3) A 26.2-kg dog is running northward at 2.59 m/s, while a 5.30-kg cat is running eastward at 3.10 m/s. Their 56.7-kg owner has the same momentum as the two pets taken together. Find the direction and magnitude of the owner's velocity.

4) Two 75.0-kg hockey players skating at 6.00 m/s collide and stick together. If the angle between their initial directions was 120

Explanation / Answer

W1=(1/2)(565)(-.02^2) changing the 2.0cm to .02m

W1= -.113J

W2=(1/2)(565)(-.02^2) changing the 2.0cm to .02m

W2= -.113J

2)0.0025 = 0.5*k*0.0048^2

k = 217N/m

b)0.0078 = 0.5*217*x^2

x = 0.8478cm

3)m1v1 +m2v2 = m3u1

56.7v = 67.858(j)+16.43(i)

v = 1.197(j) + 0.29(i)

|v| = 1.23m/s

angle = 76.38 dgrees from east towards north(north east)

4)Y: m1v1+m2v2cos120=(m1+m2)vf

75*6+75*6*cos120=(75+75)vf

vf=1.5m/s

X: m1v1+m2v2sin120=(m1+m2)vf

0+75*6*sin120=150vf

vf=2.60m/s

5)It is for when you have an elestic collision when one body is initially at rest and they have variable masses.

Vf1 = V1((m1-m2)/(m1+m2)) + V2((2m2)/(m1 + m2))

so v car = 0+17.5(2*1520)/(1520 +833)

v car = 22.6m/s

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